Does $\operatorname{Tr}(e^x(\lambda,\infty) x) =\operatorname{Tr}(px)$ imply that $e^x(\lambda,\infty) \le p \le e^x[\lambda,\infty)$? Let $H$ be a Hilbert space and $\operatorname{Tr}$ be the standard trace on $B(H)$. Let $x$ be a self-adjoint operator in $B(H)$. Let $e = e^x(\lambda,\infty)$ be the spectral projection. Assume that $\operatorname{Tr}(e x) =\operatorname{Tr}(px)$ for a projection $p$ with $\operatorname{Tr}(p)=\operatorname{Tr}(e)$. Prove that $$e^x(\lambda,\infty) \le p \le  e^x[\lambda,\infty).$$
More generally, if $p$ is a projection $\operatorname{Tr}(p)=s$ and $$\operatorname{Tr}(e )\le \operatorname{Tr}(p) \le \operatorname{Tr}( e^x[\lambda,\infty))$$
and $\operatorname{Tr}(px)= \operatorname{Tr}(ex) + \lambda (s-\operatorname{Tr}(e) )$, then do we have $$e \le p \le  e^x[\lambda,\infty).$$
 A: If you allow $\operatorname{Tr}(ex)$ to be infinite, then the implication is not true. For instance take $x=I$, $\lambda=1/2$ (so $e=I$) and $p$ any infinite projection other than the identity. 
When $x$ is positive and compact, and $\lambda>0$, the first implication is true. We have 
$$
x=\sum_j x_j e_j,
$$
where $x_1\geq x_2\geq\cdots$ and $\{e_j\}$ are pairwise orthogonal rank-one projections. Let $k$ such that $x_k=\max\{x_j:\ x_j>\lambda\}$. Then $e=\sum_{j\leq k} e_j$ and $xe=\sum_{j\geq k} x_je_j$. Thus
$$
\operatorname{Tr}(xe)=\sum_{j\leq k} x_j. 
$$
Using the orthonormal basis associated with the $\{e_j\}$ to calculate the other trace, we have 
$$
\operatorname{Tr}(px)=\sum_j x_j\,\operatorname{Tr}(pe_j).
$$
Note that $\sum_j \operatorname{Tr}(pe_j)\leq\operatorname{Tr}(p)=\operatorname{Tr}(e)=k$. Now
\begin{align}
0&=\operatorname{Tr}(xe)-\operatorname{Tr}(px)
=\sum_{j=1}^k (1-\operatorname{Tr}(pe_j))\,x_j +\sum_{j>k} x_j\operatorname{Tr}(pe_j).
\end{align}
Now all the factors in the sums are non-negative, so we have 
$$
\operatorname{Tr}(pe_j)=\begin{cases} 1,&\ 1\leq j\leq k\\ 0,&\ j>k\end{cases}.
$$
When $j\leq k$, we get 
$$
0=1-1=\operatorname{Tr}(e_j-e_jpe_j).
$$
So, being positive, we get that $e_j-e_jpe_j=0$; that is, $e_j(1-p)e_j=0$. Then $(1-p)e_j=0$ and $pe_j=e_j$. When $j>0$, $0=\operatorname{Tr}(pe_j)=\operatorname{Tr}(e_jpe_j)$, so $e_jp=0$. So $p=p_1+\sum_{j=1}^k e_j $ for some projection $p_1$ orthogonal to $\{e_j\}$; comparing traces, we get that $p=\sum_{j=1}^k e_j=e$. 
A: I might be misunderstanding your notation, but it seems like the following gives a counter-example to your first claim:
$$x =\begin{pmatrix}3 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ with $e = 1_{(2, \infty)}(x)$ and $p = 1 - e$.
A: The best answer I can find is Lemma 3.3, Remark 3.4 and Corollary 3.5   https://arxiv.org/pdf/1904.06068.pdf
